We note first that division by zero is not allowed and that for this reason \(5x-9\) may not be equal to zero and that therefore \(x={{9}\over{5}}\) is not a solution.

We now distinguish two cases, namely \(5x-9>0\) and \(5x-9<0\).
In both cases we multiply the inequality on both sides by \(5x-9\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(5x-9>0\), i.e. \(x> {{9}\over{5}}\). Then we get \(6<-8(5x-9)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(40x<66\).
Then, dvision by the coefficient of \(x\)gives \(x < {{33}\over{20}}\).
So we have the following system of inequalities: \(x> {{9}\over{5}}\,\wedge\; x < {{33}\over{20}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(5x-9<0\), i.e. \(x< {{9}\over{5}}\). Then we get \(6>-8(5x-9)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(40x>66\).
Then, division by the coefficient of \(x\) gives \(x > {{33}\over{20}}\).
So we have the following system of inequalities: \(x< {{9}\over{5}}\,\wedge\; x > {{33}\over{20}}\)
and this simplifies to \({{33}\over{20}}\lt x\lt {{9}\over{5}}\).

The solution of the original inequality is \({{33}\over{20}}\lt x\lt {{9}\over{5}}\).